There is a singularly perturbed problem:

\begin{align} &\epsilon u''(x) + u(x) - u^2(x)=0,\quad x<0<1,\\ & u(0)=1, \quad u(1)=0. \end{align}

where $\epsilon \ll 1$ is a tiny parameter. From the numerical result, it has a boundary layer at $x=1$ (The graph below is the numerical solution). I would like to find the asymptotic solution with the method of matched asymptotic expansions, but I have no idea how to deal with such a nonlinear problem.

Could anyone give me some hints? Of course, other methods or the exact solution are also available.

enter image description here

Here, I post the analytic result that I get until now (haven't finished yet). Consider the matched asymptotic expansion.

Step1: Outer Solution

Assume that the solution can be expanded in powers of $\epsilon$, namely, $$u(x,\epsilon)=u_0(x)+\epsilon u_1(x) + \mathcal{O}(\epsilon^2)$$ Substituting this into original problem and consider the $\mathcal{O}(1)$ equation, that is $\epsilon=0$, we get \begin{align} & u_0^2(x)-u_0(x)=0, \\ & u_0(0)=1. \end{align} We know that $u_0(x)=1$ is a or solution. From the graph above, it is reasonable.

Step2: Boundary Layer

Based on the numerical result, there is a boundary layer at $x=1$, so the stretching transformation we take is $$\bar x = \frac{x-1}{\delta}. $$

From the stretching transformation and the chain rule, we have that $$\frac{d}{dx}=\frac{1}{\delta}\frac{d}{d \bar x}.$$

Let the $U(\bar x)$ be the solution using the stretching transformation, then the original problem becomes \begin{align} &\frac{\epsilon}{\delta^2}U''(\bar x)+U(\bar x)-U^2(\bar x)=0, \\ &U(0)=0. \end{align} The boundary condition comes from the fact that $u(1)=U(\frac{1-1}{\delta})=U(0)=0.$ The domainant balances illustrates $\frac{\epsilon}{\delta^2}=1$, namely, $$\epsilon = \delta^2. $$ Thus, the transformed problem becomes \begin{align} &U''(\bar x)+U(\bar x)-U^2(\bar x)=0, \\ &U(0)=0. \end{align} After solving the equation above, we may get the solution that includes a parameter $c$. Then, matching the solution of boundary layer and outer solution may get the parameter $c$, and obtain the final asymptotic solution. But I have no idea how to solve the equation above.


1 Answer 1


You get a center around $u=0$ and a saddle point around $u=1$. To get at least one-sided convergence the solution has to follow the level curve of the saddle point, $$ ϵu'^2+u^2-\frac23u^2=\frac13. $$ This gives $u'(1)=\pm\frac1{\sqrt{3ϵ}}$, giving an IVP for this curve. This solution will not exactly meet $u(0)=1$, only come very close to it, with an error of the form $e^{-c/\sqrtϵ}$ which is smaller than any power of $ϵ$ asymptotically for $ϵ\to0$.

This should however be good enough for a first approximation.

  • $\begingroup$ Thanks for your answer. I drew the graph according to your answer, and the solution is $u(x)=sin(\frac{x-1}{\sqrt{3\epsilon}})$. It is indeed a good approximation. This problem, however, has several solutions. One of them is shown in the graph that I added above. So I would like to ask how can I get the asymptotic solution shown in the graph? sorry for the unclear question before. By the way, how can I get the level curve of the saddle point that you post? I study the numerical stuff and have little knowledge about the analytic one. $\endgroup$
    – ZR Tang
    Jun 15, 2021 at 12:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.